Question

cho x khác y

x²+y=y²+x

Tính giá trị biểu thức: A=( x²+y²+xy)/xy-1

Answer 1

\(x^2+y=y^2+x\Leftrightarrow x^2-y^2-\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y-1\right)=0\)

\(\Rightarrow x+y=1\)

Ta có: \(A=\frac{x^2+y^2+xy}{xy-1}\Rightarrow A+1=\frac{x^2+y^2+xy}{xy-1}+1\)

\(\Rightarrow A+1=\frac{x^2+2xy+y^2-1}{xy-1}=\frac{\left(x+y\right)^2-1}{xy-1}=\frac{0}{xy-1}=0\)

\(\Rightarrow A=-1\)